洛谷 P1816 忠诚

https://www.luogu.org/problemnew/show/1816

st表模板

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
LL m,n;
LL a[100100],d[100100][20];
int main()
{
    LL i,j,l,r,k;
    scanf("%lld%lld",&m,&n);
    for(i=1;i<=m;i++)
        scanf("%lld",&a[i]);
    for(i=1;i<=m;i++)
        d[i][0]=a[i];
    for(j=1;(1<<j)<=m;j++)//注意j和i的顺序
        for(i=1;i+(1<<j)-1<=m;i++)
            d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);
    for(i=1;i<=n;i++)
    {
        scanf("%lld%lld",&l,&r);
        k=0;
        while((1<<(k+1))<=r-l+1)    k++;
        printf("%lld ",min(d[l][k],d[r-(1<<k)+1][k]));
    }
    return 0;
}
posted @ 2018-04-04 09:09  hehe_54321  阅读(144)  评论(0编辑  收藏  举报
AmazingCounters.com